Let $D$ be a subset of $\R^n\times \R$ , $(x_0,t_0)$ a point of $D$ , and $f\colon D\to \R$ be a function.

We say that a function $x(t)$ is a solution to the Cauchy (or initial value) problem

(1)

1. $x$ is a differentiable function $x\colon I\to \R^n$ defined on a interval $I\subset \R$ ;
2. one has $(x(t),t)\in D$ for all $t\in I$ and $t_0\in I$ ;
3. one has $x(t_0)=x_0$ and $x'(t)=f(x(t),t)$ for all $t\in I$ .

We say that a solution $x\colon I\to\R^n$ is a maximal solution if it cannot be extended to a bigger interval. More precisely given any other solution $y\colon J\to \R^n$ defined on an interval $J\supset I$ and such that $y(t)=x(t)$ for all $t\in I$ , one has $I=J$ (and hence $x$ and $y$ are the same function).

We say that a solution $x\colon I\to\R^n$ is a global solution if $D\subset=\R^n \times I$ .

We say that a solution $x\colon I\to\R^n$ is unique if given any other solution $y\colon I\to\R^n$ one has $x(t)=y(t)$ for all $t\in I$ (i.e. $x$ is the unique solution defined on the interval $I$ ).

Notation

Examples

1. The function $x(t)=\log t$ defined on $I=(0,+\infty)$ is the unique maximal solution to the Cauchy problem:
   
   In this case $f(x,t)=1/t$ , $D=\{(x,t)\colon t\neq 0\}$ , $t_0=1$ , $x_0=0$ .
2. The function $x(t)=e^x$ is a global (and hence maximal), unique solution to the Cauchy problem:
   
3. Consider the Cauchy problem
   
   The function $x(t)=0$ defined on $I=\R$ is a global solution. However the function $y(t)=\sqrt{t^3}$ defined on $I=[0,+\infty)$ is also a solution and so are the functions
   
   for every $c\ge 0$ . So there are no unique solutions. Moreover $y$ is not a maximal solution.